#### Question

`f(x) = 2/x - 2/x^2, x>0`

#### Solution

\[\text { Given }: f\left( x \right) = \frac{2}{x} - \frac{2}{x^2} = 2 x^{- 1} - 2 x^{- 2} \]

\[ \Rightarrow f'\left( x \right) = - 2 x^{- 2} + 4 x^{- 3} = \frac{4}{x^3} - \frac{2}{x^2}\]

\[\text { For the local maxima or minima, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow \frac{4}{x^3} - \frac{2}{x^2} = 0\]

\[ \Rightarrow 4 - 2x = 0\]

\[ \Rightarrow x = 2\]

\[\text { Thus, x = 2 is the possible point of local maxima or local minima }. \]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{- 12}{x^4} + \frac{4}{x^3}\]

\[\text { At }x = 2: \]

\[ f''\left( 2 \right) = \frac{- 12}{16} + \frac{4}{8} = \frac{- 12 + 8}{16} = \frac{- 1}{4} < 0\]

\[\text { So, x = 2 is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( 2 \right) = \frac{2}{2} - \frac{2}{2^2} = 1 - \frac{1}{2} = \frac{1}{2}\]